If the sum of the first four numbers in a list of six consecutive even numbers is 908,
what is the sum of the last four numbers in the list?
I tried it, but got screwed up:
let 2x be one of the numbers;
list: 2x-4, 2x-2, 2x, 2x+2, 2x+4, 2x+6
sum of the first four: (2x-4) + (2x-2) + (2x) + 2x+2)
sum of last four: 2x + (2x+2) + (2x+4) + (2x+6)
=8(228) + 12
=1836...... non of the answers
Please what is the correct approach?
Let the six consecutive even numbers be \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\).
Given: \(x+(x+2)+(x+4)+(x+6)=4x+12=908\). Question: \((x+4)+(x+6)+(x+8)+(x+10)=4x+28=?\)
The way you are doing is also valid. You've just made an error in calculations, plus no need even number to be \(2x\) it can be just even \(x\).
But anyway your approach:
Sum of the first four: \((2x-4)+(2x-2)+(2x)+(2x+2)=8x-4=908\);
Sum of the last four: \(2x+(2x+2)+(2x+4)+(2x+6)=8x+12=(8x-4)+16=908+16=924\).
Hope it's clear.